OK it is possible that I have made an error earlier on in this problem, but I believe my problem is with finding the relative minimum.

Background Information:
Find the point on the parabola x+y^2=0 that is closest to the point (0, -3).

Where I believe I need help:
I found the object function to be d(y)= {(y^2-0)^2+(y+3)^2}^(1/2), where y<0.

Which I then used to do the first derivative, but I believe I either went to far or am not finding the critical points correctly some how. 1st Derivative: d'(y)= (4y^3+2y+6)/(2sqrt(y^4+(y+3)^2)), where y<0.

But then I find the C.P. to be y=-3 and y=0 which does not seem to be correct.

If anyone sees my mistake I would really appreciate the help.

Thanks,

Chris A.

May 8th 2010, 02:29 PM

mr fantastic

Quote:

Originally Posted by Cander35

OK it is possible that I have made an error earlier on in this problem, but I believe my problem is with finding the relative minimum.

Background Information:
Find the point on the parabola x+y^2=0 that is closest to the point (0, -3).

Where I believe I need help:
I found the object function to be d(y)= {(y^2-0)^2+(y+3)^2}^(1/2), where y<0.

Which I then used to do the first derivative, but I believe I either went to far or am not finding the critical points correctly some how. 1st Derivative: d'(y)= (4y^3+2y+6)/(2sqrt(y^4+(y+3)^2)), where y<0.

But then I find the C.P. to be y=-3 and y=0 which does not seem to be correct.

If anyone sees my mistake I would really appreciate the help.

Thanks,

Chris A.

How did you get y = -3 as the real solution to 4y^3 + 2y + 6 = 0? Where has